Integration By Parts (Level 2) Alright we are going to dive in and work on
various examples recall that integration by parts is particularly useful for integrands
involving products of algebraic and transcendental functions such as logarithms, exponential
functions, and trigonometric functions. Let’s try the first problem, find the integral
of x times e to the x dx. Alright we first need to notice that the integrand
contains a product of two functions specifically x and e to the x. This is a perfect candidate
to use integration by parts to find the integral. Let’s use the first form of the integration
by parts formula. Remember the first step is to make a choice
and determine which of the two functions will f(x) equal and which one will g prime of x
equal this is a critical step when applying the integration by parts formula. Keep in
mind that you want your f(x) function to be a function whose derivative f prime of x becomes
much simpler than the original f of x function. In the same manner you want your g pirme of x
function to be relatively easy to integrate. Following these steps we have that f(x) will
be represented by x because when we take its derivative we end up with a simpler function
specifically the constant 1. Likewise, we let g prime of x be equal to e to the x.
Now we need to take the derivative of f of x so the derivative of x is just equal to 1,
next we need to integrate g prime of x to obtain our g of x integrating e to the x
results in the same function e to the x. Now that we have our 4 functions, f of x f
prime of x, g of x, and g prime of x. We substitute them into the integration by parts formula
as follows, finally we take the integral of e to the x which is equal to e to x plus a
constant so the final answer is equal to, x times e to the x minus e to the x plus C.
Alright now let’s work on the same example but this time lets use the second form of
the integration by parts formula this is usually thought in a typical college level calculus
course. The first step is the same we need to decide
what function u will equal to and what function will dv equal to. Remember you want u to be
a function that becomes simpler when you take its derivative, and let dv be a function for
which you can easily find its integral. With that said we let u equal x and we let dv equal
e to the x dx make sure you include the differential dx when selecting your dv.
This is an important detail.
So now we take the derivative of u with respect to x and solve for the differential du doing
that we obtain that du is equal to dx, to obtain v we take the integral of dv or
in this case the integral of e to the x dx doing that we get that v is equal to e to
the x. Now that we have u, du, v and dv we substitute them into the integration by parts
formula, as follows, then we find the integral of e to the x, and doing that we see that
the final answer is equal to, x times e to the x minus e to the x plus C.
Alright from this point on I will use the second form of the integration by parts formula
specifically the u dv form since it’s the common formula that many college classes teach.
Personally I like using the first form but you should use the form that is easier for you.
At times you might have to rewrite the functions in the integrand in order to
apply the integration by parts formula like in the following example, find the integral
of the natural log of x over x squared dx. Notice that we don’t have a product of two
functions we actually have a quotient, but we can actually rewrite the function in the
denominator by using a particular exponent rule we can rewrite the rational
function 1 over x squared as x raised to the power of -2, and now this integral can be
rewritten as x to the negative 2 times the natural logarithm of x. notice that we actually
have a product of two functions, so now we can go ahead and apply the integration by
parts technique to find the indefinite integral. So the first step is to assign a function
for u and one for dv. Remember we want u to be a function that becomes simpler when we
differentiate it, and we want dv to be a function that can be readily integrated. If we assign
u to be equal to x to the negative 2 and take the derivative notice that we actually end
up with higher power rational function, likewise if we assign the natural log of x as
our dv we would have to take the integral of the natural logarithm which at this point
we have no idea how to find. So these particular choices of u and dv will not work. Let’s
reverse them and let u equal to the natural log of x, and dv equal to x to the power of
negative 2. Having assigned u we take the derivative with
respect to x ,which is equal to 1 over x then we solve for the differential du and obtain
1 over x dx or written another way as x to the negative 1 dx. Likewise having assigned
dv we integrate it and obtain v which is equal to negative x to the negative 1 , Now that
we have all the required functions we substitute them into the integration by parts formula,
doing that we have the following expression the natural log of x times negative x to the
negative 1 minus the integral of negative x to the negative 1 times x to the negative
1 dx. Simplifying the expressions we obtain the following, and taking the integral of
x to the negative 2 we obtain negative x to the negative 1 + C and finally we simplify
the expression and obtain the final integral of the natural log of x over x squared to
be equal to negative the natural log of x over x minus 1 over x + C.
Ok, recall that we had no idea what the integral of the natural log of x was equal to well
that is about to change we can actually use integration by parts to find the integral
of this function. Let’s try the following example, find the
integral of the natural log of x dx. Notice that we don’t have a product of two
functions we only have one function the natural log of x, but looks can be deceiving we actually
have two functions the natural log of x and the constant function 1 so we assign u to
be equal to the natural log of x and dv equal to 1 dx or more simply as dx. So taking
the derivative of u with respect to x and solving for du we obtain 1 over x dx or written
another way as x to the negative 1 dx, and taking the integral of dv we obtain v which
is equal to just x. Substituting these functions into the integration by parts formula we obtain
the following, and finally taking the integral
of dx we obtain the final integral of the natural log of x to be equal to, x times the
natural log of x minus x plus C. This example illustrates that we can actually
use integration by parts to find the integral of a single function its just a matter of
spotting and creating a second function in this case it was the constant function 1 dx.
Alright In our next video we will find more complicated integrals that require you to
repeatedly apply the integration by parts formula.